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Lisa [10]
4 years ago
14

An atom is most stable when it has eight electrons in its outermost energy level. This is a statement of the ________ rule.

Chemistry
2 answers:
Artemon [7]4 years ago
7 0

I believe your answer would be the octet rule.

Tell me if I'm wrong.

Hope this helps.

zhannawk [14.2K]4 years ago
3 0

Answer : This is a statement of the octet rule.

Explanation:

Octet rule : It is the tendency of atoms to have eight electrons in the valence shell. When the atoms have fewer than eight electrons then they tend to react and form the more stable compounds.

Or we can say that the atoms attempt to get a noble gas electron configuration, which is eight valence electrons.

For example : As we know that the sodium has one valence electron, so if giving it up then the result in the same electron configuration as the neon and chlorine has seven valence electrons, so if it takes one it will have eight and the result in the same electronic configuration as the argon.

Hence, this is a statement of the octet rule.

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3 years ago
5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------> 1 Mg3(
professor190 [17]

Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass of Mg(OH)₂ required for the reaction

= 14.8 × 58.3197 = 863.13 g

Hope this Helps!!!

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